Problem: Divide the following complex numbers. $ \dfrac{-30+10i}{-4-2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+2i}$ $ \dfrac{-30+10i}{-4-2i} = \dfrac{-30+10i}{-4-2i} \cdot \dfrac{{-4+2i}}{{-4+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-30+10i) \cdot (-4+2i)} {(-4-2i) \cdot (-4+2i)} = \dfrac{(-30+10i) \cdot (-4+2i)} {(-4)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-30+10i) \cdot (-4+2i)} {(-4)^2 - (-2i)^2} = $ $ \dfrac{(-30+10i) \cdot (-4+2i)} {16 + 4} = $ $ \dfrac{(-30+10i) \cdot (-4+2i)} {20} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-30+10i}) \cdot ({-4+2i})} {20} = $ $ \dfrac{{-30} \cdot {(-4)} + {10} \cdot {(-4) i} + {-30} \cdot {2 i} + {10} \cdot {2 i^2}} {20} $ Evaluate each product of two numbers. $ \dfrac{120 - 40i - 60i + 20 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{120 - 40i - 60i - 20} {20} = \dfrac{100 - 100i} {20} = 5-5i $